Mock AIME I 2015 Problems/Problem 11

Solution 1

For convenience, let's use $a, b, c$ instead of $\alpha, \beta, \gamma$ . Define a polynomial $P(x)$ such that $P(x) = (x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab+ac+bc)x - abc$ . Let $j = ab + ac + bc$ and $k = -abc$ . Then, our polynomial becomes $P(x) = x^3 - (a+b+c)x^2 + jx + k$ . Note that we want to compute $-\frac{j}{k}$ .

From the given information, we know that the coefficient of the $x^2$ term is $6$ , and we also know that $P(-1) = -33$ , or in other words, $-j + k = -26$ . By Newton's Sums (since we are given $a^3 + b^3 + c^3$ ), we also find that $6j + k = 43$ . Solving this system, we find that $(j, k) \in (\frac{69}{7}, -\frac{113}{7})$ . Thus, $\frac{j}{-k} = \frac{69}{113}$ , so our final answer is $69 + 113 = \boxed{182}$ .

Solution 2

Let $\alpha = a$ , $\beta = b$ , and $\beta = c$ . Then our system becomes $a + b + c = 6$ $a^3 + b^3 + c^3 = 87$ $(a + 1)(b + 1)(c + 1) = 33$ .

$a^3 + b^3 + c^3 = 87$ $a^3 + b^3 + c^3 - 3abc = 87 - 3abc$ $(a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) = 87 - 3abc$ $(a + b + c)((a + b + c)^2 - 3(ab + ac + bc)) = 87 - 3abc$ Since $a + b + c = 6$ , this equation becomes $6(6^2 - 3(ab + ac + bc)) = 87 - 3abc$ .

$33 = (a + 1)(b + 1)(c + 1) = abc + ab + ac + bc + a + b + c + 1$ . Since $a + b + c = 6$ , this equation becomes $26 = abc + ab + ac + bc$ .

We will now use these $2$ equations to solve the problem. Let $x = abc$ , and $y = ab + ac + bc$ . Then we have $6(6^2 - 3y) = 87 - 3x$ $26 = x + y$ . Our solutions are $x = \frac{113}{7}$ and $y = \frac{69}{7}$ .

Then $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{ab + ac + bc}{abc} = \frac{69/7}{113/7} = \frac{69}{113}$ . So, $m + n = 69 + 113 = \boxed{182}$ .

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Mock AIME I 2015 Problems/Problem 11

Solution 1

Solution 2